Are These Internet Circle Problems The Same? Part 2

Challenge Problem #21

In part one we tackled the first problem. You can see the final results shown below. We will use this result to tackle problem number two, beginning now.

The area of the region shaded red is “8root5 minutes 5.21pi.”  The “8root5,” part was the area of the two triangles, or the kite.  The 5.21 pi came from the sum of the sectors.  That’s key information that will allow us to tackle problem #2!  Let’s reintroduce it now.

Here we can see that the area of the shaded region is given in two parts, a term with a radical expression, “root 3,” and a term with pi.  This is fantastic because we know the radical expression comes from the Pythagorean Theorem and the term with pi comes from the area of a sector. 

We also know that the shaded region is constructed the same way!  If we connect N to K, and also M to K, we will have another kite!  The red area is the area of the kite with the two sectors removed.

Using our pictogram from before, we can label the information we’re given in the appropriate places, as shown in Figure 10.

Let’s begin by writing an expression for the kite.  To start we will consider each half of the kite,  We will use the letters A through E again, and relabel our diagram.  There’s not a mathematical reason for this, but those are the letters I used when I made up the diagrams.

In Figure 11 you can triangle ABD.  You can see that BD is the radius of circle B and that the length AB is the hypotenuse of the triangle because segment AD is tangent to circle B, which makes AD perpendicular to the radius at the point of intersection.

As was the case with problem #1, the area of the triangle is half of the base times the height.  But, we have two identical triangles making up the region of the kite ABDE, so we will double the area of the triangle.  The base is r, for the radius, and the height is AD.  The area of the kite is rad, literally, r(AD).

Even though the circles are congruent, the sector from each circle is different.  Let’s start with the sector inside of circle B.  We will use the right triangle to find the angle, then double it to find the angle of the sector.   As we did earlier, we will use the formula for the area of a sector and to create our expression for sector B, shown in Figure 13.

Angles DAE, which we need for sector A, is supplementary with angle DBE.  You can see the calculators for angle DAE and its expression in Figure 14.

We will need to use some caution when manipulating the expression for the area of sector A, but it is not too difficult.  There are just a lot of letters and expressions floating around.  This work is shown in Figure 15.

We now have an expression for each sector.  Now let’s do the algebra required to add these two together.  The result gives us “half of pi times r-squared.”

Let’s put it all together now and see if we can find the distance between the two circles’ centers.

We are given the area is “8root3 minutes 4pi.”  Remember, the “8root3,” term is from the area of the two triangles, or the kite.  The 4pi comes from the sum of the sectors removed from the kite, leaving the red-shaded region. 

We have expressions for the area where we found that the kite is r(AD), and the sum of the sectors is half of pi times r-squared.  So we know that r(AD) = “8root3,” and that 4pi = half of pi times r-squared (Figure 17).

We can solve the equation with pi for the radius, and then substitute that value into the portion that describes the area of the kite.

Now we will substitute this into the equation for the kite.

Let’s place these values on the diagram where will see that one more application of the good old Pythagorean Theorem will take it home for us, so to speak.

As shown in Figure 18, the distance connecting the centers of the two circles is “4root2,” which is a very nice answer!

Wrap-Up:

There was a simpler way to find the length of AB. I realized this when making the video for this problem.  Did you discover it?  If so, please let me know!

Other than that, how did you do?  Did you have a nicer approach than mine, or something similar?  I’d love to hear if you had something different. 

Can you create your viral math problem now?  For extra credit, try to make one, leave it in the comments and I’ll see if I can solve it.  Maybe I’ll even write it up in a future article.

As always, if you’d like to use this problem in your classroom, please visit my website, linked here, where you can the PowerPoint presentation that I used to make the video, as well as a PDF handout for your classes.

Until next week, vaya con dios, as they say in the old movies!